An electronic current is passed through two solution (A ) `AgNO_(3)` and (B) a solution of 10g of blue vitriol in 500 mL of water using platinum electrodes separately. After 30 min it was found that 1.307g Ag was deposited . What is the concentration of `Cu^(2+)` after electrolysis?

Explanation

Amount of current required to deposite 1.307 g of Ag is `:` from formula w = Zit
`1.307 = ( 107 xx 30 xx 60)/( 96500)`
i = 0.649A
now same amount of current is required for deposition of copper,
`:.` amount of `Cu^(2+)` deposited
`:.` W = Zit
`w = ( 63.5 )/( 2) xx ( 30 xx 60 xx 0.649)/( 96500)`
`w = 0.384g`
The concentration of blue vitriol solution is given by
`= ("Number of moles")/( "Volume of solution in L ") = ( 159.6g "mol"^(-1))/( 500 xx 10^(-3) L ) = 0.125M`
Concentration of `CuSO_(4)` that gets deposited
`= ((0.384g)/(159.6 g "mol"^(-1)))/(500 xx 10^(-3) L) = 4.8 xx 10^(-3) M`
`:. ` Concentration of `Cu^(2+)` after electrolysis
= 0.125M `- 4.8 xx 10^(-3) M` = 0.120M